The graph shows the function g(x) for a restricted domain.

On a coordinate plane, a function starts at (negative 4, 0) and then curves through the y-axis at (0, 1.5) and then goes through point (4, 2).

Which is the function g(x) for a restricted domain?

g(x) = Negative RootIndex 3 StartRoot x minus 4 EndRoot; x greater-than-or-equal-to –4
g(x) = Negative RootIndex 3 StartRoot x 4 EndRoot + 4; x greater-than-or-equal-to 0
g(x) = Negative RootIndex 3 StartRoot x + 4 EndRoot; x greater-than-or-equal-to –4
g(x) = Negative RootIndex 3 StartRoot x EndRoot minus 4; x greater-than-or-equal-to 0

Since opening night, attendance at play a has increased steadily, while attendance at play b first rose and then fell. equations modeling the daily attendance y at each play are shown below, where x is the number of days since opening night. on what day(s) was the attendance the same at both plays? what was the attendance? play a: y = 8x + 191 play b: y = -x^2 + 26x + 126 a. the attendance was never the same at both plays. b. the attendance was the same on day 5. the attendance was 231 at both plays on that day. c. the attendance was the same on day 13. the attendance was 295 at both plays on that day. d. the attendance was the same on days 5 and 13. the attendance at both plays on those days was 231 and 295 respectively.

What is the slope of the line that passes through the points (–9, 2) and (0, 4)? the slope of the line is

Write the component forms of vectors u and v, shown in the graph, and find v − 2u. u= (< -3, -2> , < -3, -1> , < -2, -2> , < -2, -1> ) v= (< -5, 1> , -4, 0> , < 0, -4> , < 1, -5> ) v-2u= (< 5, 3> , < 0, 4> , < 4, 0> , < 5, -3>